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# m05 Q5 simple math help please

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28 Jul 2013, 22:12
P@=P1/P−1

Lets take p=3
3@ = 3/2

(P@)@= (3/2)@= (3/2)/(1/2)=3
Hence option C
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28 Jul 2013, 22:29
vaishnogmat wrote:
If P@=P1/P−1, what is the value of P@@?

(A) P/P−1

(B) 1/P

(C) P

(D) 2−P

(E) P−1

[Reveal] Spoiler:
C

Merging similar topics.
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29 Jul 2013, 09:21
fameatop wrote:
P@=P1/P−1

Lets take p=3
3@ = 3/2

(P@)@= (3/2)@= (3/2)/(1/2)=3
Hence option C

Thanks for your explanation. I understand the part where you add value for P@, but when you go from P@ to (P@)@ that's where I get stuck.

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06 Sep 2013, 03:20
dzyubam wrote:
I'll try to make this clearer by using our math syntax. See the short intro here:

Given $$P# = \frac{P}{P-1}$$, need to find $$P##$$ - original problem
We might simplify this problem by saying that this '#' sign is an operation that replaces each $$P$$ with $$\frac{P}{P-1}$$. $$P##$$ is found by replacing all the $$P$$s with $$\frac{P}{P-1}$$ in this expression:
$$\frac{P}{P-1}$$, which equals $$P#$$

This is how it looks after these replacements:

$$\frac{\frac{P}{P-1}}{\frac{P}{P-1}-1}$$

$$\frac{\frac{P}{P-1}}{\frac{P}{P-1}-\frac{P-1}{P-1}}=$$

$$\frac{\frac{P}{P-1}}{\frac{P-P+1}{P-1}}=$$

$$\frac{\frac{P}{P-1}}{\frac{1}{P-1}}=$$

$$\frac{P}{P-1}*\frac{P-1}{1}=$$

$$P$$

Hope this clears the doubts.

I think that's a baldy written question. Why not P## be a multiplication? Similarly to other questions like:
What is $$(2^3)(4^5)(6^7)$$ ? Which straightly implies multiplication of the 3 numbers (as opposed to ask for a number where the 1st digit(s) would be 2^3, the next ones 4^5 and the last ones 6^7.
(Actually I got confused and ended up multiplying # * #).

In m opinion it should be written as a function (P(P#)) or something.
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20 Apr 2014, 07:26
P##=(P/P-1)/(P/P-1-1)=(P/P-1)/(1/P-1)=P
Re: m05 Q5 simple math help please   [#permalink] 20 Apr 2014, 07:26

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# m05 Q5 simple math help please

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